If $G(q),H(q)$ are the functions appearing in Rogers-Ramanujan identities $$G(q)=\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(q;q)_{n}}=\prod_{n=1}^{\infty}\frac{1}{(1-q^{5n-1})(1-q^{5n-4})}$$ and $$H(q)=\sum_{n=0}^{\infty}\frac{q^{n^{2}+n}}{(q;q)_{n}}=\prod_{n=1}^{\infty}\frac{1}{(1-q^{5n-2})(1-q^{5n-3})}$$ then it is known that $q^{-1/60}G(q)$ and $q^{11/60}H(q)$ are modular forms of same weight and hence each of them must be expressible in the form $$\left(\frac{2K}{\pi}\right)^{s}A(k)$$ where $s$ is the weight of the modular form and $A(k)$ is some algebraic function of $k$. Do we know the explicit expressions for $G(q),H(q)$ in terms of $K,k$? I would want to use these expressions (coupled with modular equations) to prove Ramanujan's identities like $$G(q^{11})H(q)-q^{2}G(q)H(q^{11})=1$$
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$\begingroup$ I think such expressions are likely to be extremely unwieldy. $q^{1/5} G(q)/H(q)$ is a generator for the field of functions on $\tilde{X}(5)$, which is a degree $60$ cover of $X(1)$ and $k$ (which I presume is $\vartheta_{2}(q)^{2}/\vartheta_{3}(q)^{2}$) is a generator for the field of modular functions on $X_{0}(4)$. Thus the algebraic relation between $k$ and $R = q^{1/5} G(q)/H(q)$ has degree $60$ in $R$ and degree $3$ in $k$. I would guess the individual expressions for $G(q)$ and $H(q)$ would be worse. $\endgroup$– Jeremy RouseSep 26, 2014 at 16:55
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$\begingroup$ @JeremyRouse: Thanks for the inputs. The relation between $R$ and $k$ is definitely complicated but at least we have relation like $$\frac{1}{R} -1-R=\frac{f(-q^{1/5})}{q^{1/5}f(-q^{5})}$$ where $f(-q)=(q;q)_{\infty}$ is essentially the eta function. I don't think $G(q), H(q)$ would have any simple relation with eta function individually. I guess I will have to use the product form of $G(q),H(q)$ to prove the Ramanujan identities. $\endgroup$– Paramanand SinghSep 27, 2014 at 2:46
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$\begingroup$ @ParamanandSingh: You may like this related post. $\endgroup$– Tito Piezas IIIJul 7, 2015 at 4:51
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$\begingroup$ @TitoPiezasIII: I am reading your post. It seems pretty interesting. $\endgroup$– Paramanand SinghJul 7, 2015 at 5:27
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